Three coins are tossed once. Find the probability of getting (i) 3 heads (ii) 2 heads (iii) atleast 2 heads (i

Exercise 14.2 · Q8

Q. Three coins are tossed once. Find the probability of getting

(i) 3 heads

(ii) 2 heads

(iii) atleast 2 heads

(iv) atmost 2 heads

(v) no head

(vi) 3 tails

(vii) exactly two tails

(viii) no tail

(ix) atmost two tails.

Himachal HpboseTextbookSubjective· 3mImportance★★★★★est

25% · 23/93 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Sample Space Outcomes

Sample Space Outcomes: The Complete Picture of "What Could Happen?"

Imagine you're about to flip a coin. Before it lands, you know two things are possible: heads or tails. That set of all possible results — {Heads, Tails} — is your sample space. Every single thing that could happen, listed out completely.

That's the core idea. Let's build from there.

The Intuition First

When you roll a six-sided die, you don't know which number will come up. But you do know the full list of possibilities: 1, 2, 3, 4, 5, or 6. That list is the sample space. Each individual possibility — like rolling a 4 — is an outcome (also called a sample point).

Think of it as the "universe" of your experiment. Nothing outside this set can happen. If you're drawing a card from a standard deck, the sample space is all 52 cards. If you're checking the weather tomorrow (sunny, rainy, cloudy), those three options form the sample space.

Note

The sample space is exhaustive — it covers every possible result — and its outcomes are mutually exclusive (no two can happen at the same time in a single trial).

The Precise Statement

In probability theory, an experiment is any process with uncertain outcomes (rolling a die, drawing a card, measuring rainfall). The sample space, denoted by $S$ or $\Omega$ , is the set of all possible outcomes of that experiment.

Each element of $S$ is called an outcome (or sample point). For a fair coin toss:

S = \{H, T\}

For a single die roll:

S = \{1, 2, 3, 4, 5, 6\}

For the number of heads in two coin tosses:

S = \{0, 1, 2\}

S = \{\text{all possible outcomes of the experiment}\}

Two Important Flavors

Discrete sample space — outcomes can be listed (countable). Coin tosses, die rolls, number of customers in a queue.

Continuous sample space — outcomes form a range (uncountable). The exact time a bus arrives (any real number between 0 and 60 minutes), the temperature at noon, the height of a randomly selected student.

For continuous cases, we write intervals: $S = \{x \mid 0 \leq x \leq 60\}$ for bus arrival time in minutes.

Why This Matters

Every probability question starts with the sample space. If you don't know what could happen, you can't calculate the chance of a specific event. The sample space is the foundation — get it right, and everything else follows. …

Start your 14-day free trial to unlock the full solution →

Concept: Sample Space Outcomes – When three coins are tossed, each coin has two outcomes (H or T), so total outcomes = $2^3 = 8$ . The sample space is:

S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}

Reasoning:

Count favourable outcomes for each event.
Probability = $\frac{\text{Favourable outcomes}}{\text{Total outcomes}} = \frac{\text{count}}{8}$ .

Event	Favourable outcomes	Count	Probability
(i) 3 heads	HHH	1	$\frac{1}{8}$
(ii) 2 heads	HHT, HTH, THH	3	$\frac{3}{8}$
(iii) atleast 2 heads	HHH, HHT, HTH, THH	4	$\frac{4}{8} = \frac{1}{2}$
(iv) atmost 2 heads	all except HHH	7	$\frac{7}{8}$
(v) no head	TTT	1	$\frac{1}{8}$
(vi) 3 tails	TTT	1	$\frac{1}{8}$

Start your 14-day free trial to unlock the full solution →

The sample space for tossing three coins has 8 equally likely outcomes. Using this, each probability is simply the count of favourable outcomes divided by 8. The answers range from $\frac{1}{8}$ to $\frac{7}{8}$ .

When you toss three coins, each coin lands either Head (H) or Tail (T). The key insight: every coin is independent, so the total number of possible results is $2 \times 2 \times 2 = 8$ . These 8 outcomes are equally likely — no outcome is more probable than another. This makes probability just a matter of counting.

The sample space is:

S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}

Now we go through each part.

3 heads

Only one outcome: $HHH$ .

Probability $= \frac{1}{8}$ .
2 heads

Outcomes with exactly two H's: $HHT, HTH, THH$ — that's 3.

Probability $= \frac{3}{8}$ .
At least 2 heads

This means 2 heads or 3 heads. Count: 3 (two heads) + 1 (three heads) = 4.

Probability $= \frac{4}{8} = \frac{1}{2}$ .
At most 2 heads

This means 0, 1, or 2 heads. Easier: total outcomes (8) minus the one with 3 heads = 7.

Probability $= \frac{7}{8}$ .
No head

That means all tails: $TTT$ — only 1 outcome.

Probability $= \frac{1}{8}$ .
3 tails

Same as no head: $TTT$ — 1 outcome.

Probability $= \frac{1}{8}$ .
Exactly two tails

Outcomes with exactly two T's: $HTT, THT, TTH$ — 3 outcomes.

Probability $= \frac{3}{8}$ .
No tail

That means all heads: $HHH$ — 1 outcome.

Probability $= \frac{1}{8}$ .
At most two tails …

Start your 14-day free trial to unlock the full solution →

←Previous question · 14.2.4Q7 — A fair coin is tossed four times, and a person win Re 1 for each head …→Next question · 14.2.4Q9 — If (2)/(11) is the probability of an event A, what is the probability …

Q.Three coins are tossed once. Find the probability of getting

🎓Unlock everything free for 14 days

Unlock everything free for 14 days